Matrices $n \times n$ on complex field. Compute Jordan form of operator $X \mapsto AXA$: $$ A = \begin{bmatrix} 0 & 1 & & \\ & 0 & \ddots & \\ & & \ddots & 1 \\ & & & 0 \end{bmatrix} $$ A is nilpotent Jordan block


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  • $\begingroup$ do you mean $X \mapsto AXA$? Also, what have you tried? $\endgroup$ – peek-a-boo Jul 22 at 1:37

Hint: The Jordan form of the map $T(X) = AXA$ can be deduced using (only) the following pieces of information:

  • $T$ is a linear map on a space with dimension $n^2$
  • $T^n = 0$
  • More generally, $\operatorname{rank}(T^{k}) = (n-k)^2$, $k = 1,\dots,n$

Another approach: using the vectorization operator, we can conclude that the matrix of your transformation (relative to a certain basis of $\Bbb C^{n \times n}$) is $A^T \otimes A$, where $\otimes$ denotes the Kronecker product. This matrix is "almost" in Jordan normal form.

My original answer to this question, where I incorrectly implied that the Jordan form of $T$ would be $I_n \otimes A$ was incorrect; the current version is correct.

To see that $\operatorname{rank}(T^{k}) = (n-k)^2$, $k = 1,\dots,n$ holds, it suffices to make the following observation. The domain $\Bbb C^{n \times n}$ is spanned by elements of the form $uv^T$ with $u,v \in \Bbb C^n$. Thus, the image of $T^k$ is spanned by elements of the form $$ T^k(uv^T) = (A^k u)(v^T A^k) = (A^n u)((A^T)^kv)^T. $$ Thus, the image of $T^k$ is spanned by the matrices $xy^T$ where $x,y$ are in the images of $A^k$ and $(A^T)^k$ respectively. Because the image of $A^k$ and $(A^T)^k$ each have dimension $n-k$, we may conclude that the image of $T^k$ has dimension $(n-k)^2$.

Since $T$ is nilpotent, every Jordan block in the Jordan form is a block associated with $0$. More specifically, we may use the above observation to conclude that $T$ has $1$ block of size $n$, and $2$ blocks of size $k$ for $k = 1,\dots,n-1$.

  • 1
    $\begingroup$ Try to prove that the second property is true, you should find the proof straightforward. If you spend more than 15 minutes on it, let me know and I’ll add an explanation $\endgroup$ – Omnomnomnom Jul 22 at 17:54
  • $\begingroup$ My original answer was wrong, sorry about that. See my latest edit. $\endgroup$ – Omnomnomnom Jul 25 at 19:38

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